Prove that (1)Let a,b ? R. Then a-b is a factor of a^n-b^n for all n ? N. That i

Question

Prove that (1)Let a,b ? R. Then a-b is a factor of a^n-b^n for all n ? N. That is, for each n ? N, we have a^n-b^n = (a-b)c for some c, where c is a polynomial of two variables a,b. (2)2 ?... ? 2 = 2^n ? (n+1)! = (n+1) ? n ?...? 2 ? 1 for all n ? N. (3)For each r ? R{1}, we have 1+r+r^2+...+r^n = [1-r^(n+1)] / (1-r) for all n ? N. (4)For each x ? R, x > -1, we have (1+x)^n ? 1 + nx for all n ? N. (5)Let n ? N and a1,a2,...,an+1 ? R. Then max {a1,a2,,...,an+1} = ai for some 1? i ? n+1 min {a1,a2,...,an+1} = aj for some 1 ? j ? n+1

Explanation / Answer

first we check for n = 1. a^n - b^n = a-b this is divisible by a-b so the statement is true for n=1. let us assume that it is true for some k where k is a natural number. i.e. a^k - b^k = p*(a-b) where p is a polynomial in terms of a and b ............... 1 now for n = k+1, a^(k+1) - b^(k+1) = a^k*a - b^k*b from 1, a^k = b^k + p*(a-b) so a^(k+1) - b^(k+1) = [b^k + p*(a-b)]*a - b^k*b = b^k*a + ap*(a-b) - b^k*b

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